$NH _{2} NH _{2}+ H _{2} O \rightleftharpoons NH _{2} NH _{3}^{+}+ OH ^{-}$

From the $pH$ we can calculate the hydrogen ion concentration. Knowing hydrogen ion concentration and the ionic product of water we can calculate the concentration of hydroxyl ions. Thus we have: 

$\left.| H ^{+}\right]=$ antilog $(- pH )$

$=$ antilog $(-9.7)=1.67 \times 10^{-10}$

$\left[ OH ^{-}\right]=K_{ w } /\left[ H ^{+}\right] =1 \times 10^{-14} / 1.67 \times 10^{-10} $

$=5.98 \times 10^{-5}$

The concentration of the corresponding hydrazinium ion is also the same as that of hydroxyl ion. The concentration of both these ions is very small so the concentration of the undissociated base can be taken equal to $0.004 \,M$ Thus,

$K_{ b }=\left[ NH _{2} NH _{3}^{+}\right]\left[ OH ^{\top}\right] /\left[ NH _{2} NH _{2}\right]$

$=\left(5.98 \times 10^{-5}\right)^{2} / 0.004=8.96 \times 10^{-7}$

$p K_{ b }=-\log K_{ b }=-\log \left(8.96 \times 10^{-7}\right)=6.04$